59.88 Base change for higher direct images
This section is the analogue of Section 59.87 for higher direct images. This section is preliminary and should be skipped on a first reading.
Lemma 59.88.2. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have $BC(f, n, q - 1)$. Then for every commutative diagram
\[ \xymatrix{ X \ar[d]_ f & X' \ar[l] \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e \\ S & S' \ar[l] & T \ar[l]_ g } \]
with $X' = X \times _ S S'$ and $Y = X' \times _{S'} T$ and $g$ quasi-compact and quasi-separated, and every abelian sheaf $\mathcal{F}$ on $T_{\acute{e}tale}$ annihilated by $n$
the base change map $(f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F}$ is injective,
if $\mathcal{F} \subset \mathcal{G}$ where $\mathcal{G}$ on $T_{\acute{e}tale}$ is annihilated by $n$, then
\[ \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{G}\to R^ qh_*e^{-1}\mathcal{G} \right) \]
if in (2) the sheaf $\mathcal{G}$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules, then
\[ \mathop{\mathrm{Coker}}\left((f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset R^ qh_*e^{-1}\mathcal{G} \]
Proof.
Choose a short exact sequence $0 \to \mathcal{F} \to \mathcal{I} \to \mathcal{Q} \to 0$ where $\mathcal{I}$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules. Consider the induced diagram
\[ \xymatrix{ (f')^{-1}R^{q - 1}g_*\mathcal{I} \ar[d]_{\cong } \ar[r] & (f')^{-1}R^{q - 1}g_*\mathcal{Q} \ar[d]_{\cong } \ar[r] & (f')^{-1}R^ qg_*\mathcal{F} \ar[d] \ar[r] & 0 \ar[d] \\ R^{q - 1}h_*e^{-1}\mathcal{I} \ar[r] & R^{q - 1}h_*e^{-1}\mathcal{Q} \ar[r] & R^ qh_*e^{-1}\mathcal{F} \ar[r] & R^ qh_*e^{-1}\mathcal{I} } \]
with exact rows. We have the zero in the right upper corner as $\mathcal{I}$ is injective. The left two vertical arrows are isomorphisms by $BC(f, n, q - 1)$. We conclude that part (1) holds. The above also shows that
\[ \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset R^ qh_*e^{-1}\mathcal{I} \]
hence part (3) holds. To prove (2) choose $\mathcal{F} \subset \mathcal{G} \subset \mathcal{I}$.
$\square$
Lemma 59.88.3. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have $BC(f, n, q - 1)$. Consider commutative diagrams
\[ \vcenter { \xymatrix{ X \ar[d]_ f & X' \ar[d]_{f'} \ar[l] & Y \ar[l]^ h \ar[d]^ e & Y' \ar[l]^{\pi '} \ar[d]^{e'} \\ S & S' \ar[l] & T \ar[l]_ g & T' \ar[l]_\pi } } \quad \text{and}\quad \vcenter { \xymatrix{ X' \ar[d]_{f'} & & Y' \ar[ll]^{h' = h \circ \pi '} \ar[d]^{e'} \\ S' & & T' \ar[ll]_{g' = g \circ \pi } } } \]
where all squares are cartesian, $g$ quasi-compact and quasi-separated, and $\pi $ is integral surjective. Let $\mathcal{F}$ be an abelian sheaf on $T_{\acute{e}tale}$ annihilated by $n$ and set $\mathcal{F}' = \pi ^{-1}\mathcal{F}$. If the base change map
\[ (f')^{-1}R^ qg'_*\mathcal{F}' \longrightarrow R^ qh'_*(e')^{-1}\mathcal{F}' \]
is an isomorphism, then the base change map $(f')^{-1}R^ qg_*\mathcal{F} \to R^ qh_*e^{-1}\mathcal{F}$ is an isomorphism.
Proof.
Observe that $\mathcal{F} \to \pi _*\pi ^{-1}\mathcal{F}'$ is injective as $\pi $ is surjective (check on stalks). Thus by Lemma 59.88.2 we see that it suffices to show that the base change map
\[ (f')^{-1}R^ qg_*\pi _*\mathcal{F}' \longrightarrow R^ qh_*e^{-1}\pi _*\mathcal{F}' \]
is an isomorphism. This follows from the assumption because we have $R^ qg_*\pi _*\mathcal{F}' = R^ qg'_*\mathcal{F}'$, we have $e^{-1}\pi _*\mathcal{F}' =\pi '_*(e')^{-1}\mathcal{F}'$, and we have $R^ qh_*\pi '_*(e')^{-1}\mathcal{F}' = R^ qh'_*(e')^{-1}\mathcal{F}'$. This follows from Lemmas 59.55.4 and 59.43.5 and the relative leray spectral sequence (Cohomology on Sites, Lemma 21.14.7).
$\square$
Lemma 59.88.4. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have $BC(f, n, q - 1)$. Consider commutative diagrams
\[ \vcenter { \xymatrix{ X \ar[d]_ f & X' \ar[d]_{f'} \ar[l] & X'' \ar[l]^{\pi '} \ar[d]_{f''} & Y \ar[l]^{h'} \ar[d]^ e \\ S & S' \ar[l] & S'' \ar[l]_\pi & T \ar[l]_{g'} } } \quad \text{and}\quad \vcenter { \xymatrix{ X' \ar[d]_{f'} & & Y \ar[ll]^{h = h' \circ \pi '} \ar[d]^ e \\ S' & & T \ar[ll]_{g = g' \circ \pi } } } \]
where all squares are cartesian, $g'$ quasi-compact and quasi-separated, and $\pi $ is integral. Let $\mathcal{F}$ be an abelian sheaf on $T_{\acute{e}tale}$ annihilated by $n$. If the base change map
\[ (f')^{-1}R^ qg_*\mathcal{F} \longrightarrow R^ qh_*e^{-1}\mathcal{F} \]
is an isomorphism, then the base change map $(f'')^{-1}R^ qg'_*\mathcal{F} \to R^ qh'_*e^{-1}\mathcal{F}$ is an isomorphism.
Proof.
Since $\pi $ and $\pi '$ are integral we have $R\pi _* = \pi _*$ and $R\pi '_* = \pi '_*$, see Lemma 59.43.5. We also have $(f')^{-1}\pi _* = \pi '_*(f'')^{-1}$. Thus we see that $\pi '_*(f'')^{-1}R^ qg'_*\mathcal{F} = (f')^{-1}R^ qg_*\mathcal{F}$ and $\pi '_*R^ qh'_*e^{-1}\mathcal{F} = R^ qh_*e^{-1}\mathcal{F}$. Thus the assumption means that our map becomes an isomorphism after applying the functor $\pi '_*$. Hence we see that it is an isomorphism by Lemma 59.43.5.
$\square$
Lemma 59.88.5. Let $T$ be a quasi-compact and quasi-separated scheme. Let $P$ be a property for quasi-compact and quasi-separated schemes over $T$. Assume
If $T'' \to T'$ is a thickening of quasi-compact and quasi-separated schemes over $T$, then $P(T'')$ if and only if $P(T')$.
If $T' = \mathop{\mathrm{lim}}\nolimits T_ i$ is a limit of an inverse system of quasi-compact and quasi-separated schemes over $T$ with affine transition morphisms and $P(T_ i)$ holds for all $i$, then $P(T')$ holds.
If $Z \subset T'$ is a closed subscheme with quasi-compact complement $V \subset T'$ and $P(T')$ holds, then either $P(V)$ or $P(Z)$ holds.
Then $P(T)$ implies $P(\mathop{\mathrm{Spec}}(K))$ for some morphism $\mathop{\mathrm{Spec}}(K) \to T$ where $K$ is a field.
Proof.
Consider the set $\mathfrak T$ of closed subschemes $T' \subset T$ such that $P(T')$. By assumption (2) this set has a minimal element, say $T'$. By assumption (1) we see that $T'$ is reduced. Let $\eta \in T'$ be the generic point of an irreducible component of $T'$. Then $\eta = \mathop{\mathrm{Spec}}(K)$ for some field $K$ and $\eta = \mathop{\mathrm{lim}}\nolimits V$ where the limit is over the affine open subschemes $V \subset T'$ containing $\eta $. By assumption (3) and the minimality of $T'$ we see that $P(V)$ holds for all these $V$. Hence $P(\eta )$ by (2) and the proof is complete.
$\square$
Lemma 59.88.6. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have that $BC(f, n, q - 1)$ is true, but $BC(f, n, q)$ is not. Then there exist a commutative diagram
\[ \xymatrix{ X \ar[d]_ f & X' \ar[d]_{f'} \ar[l] & Y \ar[l]^ h \ar[d]^ e \\ S & S' \ar[l] & \mathop{\mathrm{Spec}}(K) \ar[l]_ g } \]
where $X' = X \times _ S S'$, $Y = X' \times _{S'} \mathop{\mathrm{Spec}}(K)$, $K$ is a field, and $\mathcal{F}$ is an abelian sheaf on $\mathop{\mathrm{Spec}}(K)$ annihilated by $n$ such that $(f')^{-1}R^ qg_*\mathcal{F} \to R^ qh_*e^{-1}\mathcal{F}$ is not an isomorphism.
Proof.
Choose a commutative diagram
\[ \xymatrix{ X \ar[d]_ f & X' \ar[l] \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e \\ S & S' \ar[l] & T \ar[l]_ g } \]
with $X' = X \times _ S S'$ and $Y = X' \times _{S'} T$ and $g$ quasi-compact and quasi-separated, and an abelian sheaf $\mathcal{F}$ on $T_{\acute{e}tale}$ annihilated by $n$ such that the base change map $(f')^{-1}R^ qg_*\mathcal{F} \to R^ qh_*e^{-1}\mathcal{F}$ is not an isomorphism. Of course we may and do replace $S'$ by an affine open of $S'$; this implies that $T$ is quasi-compact and quasi-separated. By Lemma 59.88.2 we see $(f')^{-1}R^ qg_*\mathcal{F} \to R^ qh_*e^{-1}\mathcal{F}$ is injective. Pick a geometric point $\overline{x}$ of $X'$ and an element $\xi $ of $(R^ qh_*q^{-1}\mathcal{F})_{\overline{x}}$ which is not in the image of the map $((f')^{-1}R^ qg_*\mathcal{F})_{\overline{x}} \to (R^ qh_*e^{-1}\mathcal{F})_{\overline{x}}$.
Consider a morphism $\pi : T' \to T$ with $T'$ quasi-compact and quasi-separated and denote $\mathcal{F}' = \pi ^{-1}\mathcal{F}$. Denote $\pi ' : Y' = Y \times _ T T' \to Y$ the base change of $\pi $ and $e' : Y' \to T'$ the base change of $e$. Picture
\[ \vcenter { \xymatrix{ X' \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e & Y' \ar[l]^{\pi '} \ar[d]^{e'} \\ S' & T \ar[l]_ g & T' \ar[l]_\pi } } \quad \text{and}\quad \vcenter { \xymatrix{ X' \ar[d]_{f'} & & Y' \ar[ll]^{h' = h \circ \pi '} \ar[d]^{e'} \\ S' & & T' \ar[ll]_{g' = g \circ \pi } } } \]
Using pullback maps we obtain a canonical commutative diagram
\[ \xymatrix{ (f')^{-1}R^ qg_*\mathcal{F} \ar[r] \ar[d] & (f')^{-1}R^ qg'_*\mathcal{F}' \ar[d] \\ R^ qh_*e^{-1}\mathcal{F} \ar[r] & R^ qh'_*(e')^{-1}\mathcal{F}' } \]
of abelian sheaves on $X'$. Let $P(T')$ be the property
We claim that hypotheses (1), (2), and (3) of Lemma 59.88.5 hold for $P$ which proves our lemma.
Condition (1) of Lemma 59.88.5 holds for $P$ because the étale topology of a scheme and a thickening of the scheme is the same. See Proposition 59.45.4.
Suppose that $I$ is a directed set and that $T_ i$ is an inverse system over $I$ of quasi-compact and quasi-separated schemes over $T$ with affine transition morphisms. Set $T' = \mathop{\mathrm{lim}}\nolimits T_ i$. Denote $\mathcal{F}'$ and $\mathcal{F}_ i$ the pullback of $\mathcal{F}$ to $T'$, resp. $T_ i$. Consider the diagrams
\[ \vcenter { \xymatrix{ X \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e & Y_ i \ar[l]^{\pi _ i'} \ar[d]^{e_ i} \\ S & T \ar[l]_ g & T_ i \ar[l]_{\pi _ i} } } \quad \text{and}\quad \vcenter { \xymatrix{ X \ar[d]_{f'} & & Y_ i \ar[ll]^{h_ i = h \circ \pi _ i'} \ar[d]^{e_ i} \\ S & & T_ i \ar[ll]_{g_ i = g \circ \pi _ i} } } \]
as in the previous paragraph. It is clear that $\mathcal{F}'$ on $T'$ is the colimit of the pullbacks of $\mathcal{F}_ i$ to $T'$ and that $(e')^{-1}\mathcal{F}'$ is the colimit of the pullbacks of $e_ i^{-1}\mathcal{F}_ i$ to $Y'$. By Lemma 59.51.8 we have
\[ R^ qh'_*(e')^{-1}\mathcal{F}' = \mathop{\mathrm{colim}}\nolimits R^ qh_{i, *}e_ i^{-1}\mathcal{F}_ i \quad \text{and}\quad (f')^{-1}R^ qg'_*\mathcal{F}' = \mathop{\mathrm{colim}}\nolimits (f')^{-1}R^ qg_{i, *}\mathcal{F}_ i \]
It follows that if $P(T_ i)$ is true for all $i$, then $P(T')$ holds. Thus condition (2) of Lemma 59.88.5 holds for $P$.
The most interesting is condition (3) of Lemma 59.88.5. Assume $T'$ is a quasi-compact and quasi-separated scheme over $T$ such that $P(T')$ is true. Let $Z \subset T'$ be a closed subscheme with complement $V \subset T'$ quasi-compact. Consider the diagram
\[ \xymatrix{ Y' \times _{T'} Z \ar[d]_{e_ Z} \ar[r]_{i'} & Y' \ar[d]_{e'} & Y' \times _{T'} V \ar[l]^{j'} \ar[d]^{e_ V} \\ Z \ar[r]^ i & T' & V \ar[l]_ j } \]
Choose an injective map $j^{-1}\mathcal{F}' \to \mathcal{J}$ where $\mathcal{J}$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules on $V$. Looking at stalks we see that the map
\[ \mathcal{F}' \to \mathcal{G} = j_*\mathcal{J} \oplus i_*i^{-1}\mathcal{F}' \]
is injective. Thus $\xi '$ maps to a nonzero element of
\begin{align*} & \mathop{\mathrm{Coker}}\left( ((f')^{-1}R^ qg'_*\mathcal{G})_{\overline{x}} \to (R^ qh'_*(e')^{-1}\mathcal{G})_{\overline{x}} \right) = \\ & \mathop{\mathrm{Coker}}\left( ((f')^{-1}R^ qg'_*j_*\mathcal{J})_{\overline{x}} \to (R^ qh'_*(e')^{-1}j_*\mathcal{J})_{\overline{x}} \right) \oplus \\ & \mathop{\mathrm{Coker}}\left( ((f')^{-1}R^ qg'_*i_*i^{-1}\mathcal{F}')_{\overline{x}} \to (R^ qh'_*(e')^{-1}i_*i^{-1}\mathcal{F}')_{\overline{x}} \right) \end{align*}
by part (2) of Lemma 59.88.2. If $\xi '$ does not map to zero in the second summand, then we use
\[ (f')^{-1}R^ qg'_*i_*i^{-1}\mathcal{F}' = (f')^{-1}R^ q(g' \circ i)_*i^{-1}\mathcal{F}' \]
(because $Ri_* = i_*$ by Proposition 59.55.2) and
\[ R^ qh'_*(e')^{-1}i_*i^{-1}\mathcal{F} = R^ qh'_*i'_*e_ Z^{-1}i^{-1}\mathcal{F} = R^ q(h' \circ i')_*e_ Z^{-1}i^{-1}\mathcal{F}' \]
(first equality by Lemma 59.55.3 and the second because $Ri'_* = i'_*$ by Proposition 59.55.2) to we see that we have $P(Z)$. Finally, suppose $\xi '$ does not map to zero in the first summand. We have
\[ (e')^{-1}j_*\mathcal{J} = j'_*e_ V^{-1}\mathcal{J} \quad \text{and}\quad R^ aj'_*e_ V^{-1}\mathcal{J} = 0, \quad a = 1, \ldots , q - 1 \]
by $BC(f, n, q - 1)$ applied to the diagram
\[ \xymatrix{ X \ar[d]_ f & Y' \ar[l] \ar[d]_{e'} & Y \ar[l]^{j'} \ar[d]^{e_ V} \\ S & T' \ar[l] & V \ar[l]_ j } \]
and the fact that $\mathcal{J}$ is injective. By the relative Leray spectral sequence for $h' \circ j'$ (Cohomology on Sites, Lemma 21.14.7) we deduce that
\[ R^ qh'_*(e')^{-1}j_*\mathcal{J} = R^ qh'_*j'_*e_ V^{-1}\mathcal{J} \longrightarrow R^ q(h' \circ j')_* e_ V^{-1}\mathcal{J} \]
is injective. Thus $\xi $ maps to a nonzero element of $(R^ q(h' \circ j')_* e_ V^{-1}\mathcal{J})_{\overline{x}}$. Applying part (3) of Lemma 59.88.2 to the injection $j^{-1}\mathcal{F}' \to \mathcal{J}$ we conclude that $P(V)$ holds.
$\square$
Lemma 59.88.7. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have that $BC(f, n, q - 1)$ is true, but $BC(f, n, q)$ is not. Then there exist a commutative diagram
\[ \xymatrix{ X \ar[d]_ f & X' \ar[d] \ar[l] & Y \ar[l]^ h \ar[d] \\ S & S' \ar[l] & \mathop{\mathrm{Spec}}(K) \ar[l] } \]
with both squares cartesian, where
$S'$ is affine, integral, and normal with algebraically closed function field,
$K$ is algebraically closed and $\mathop{\mathrm{Spec}}(K) \to S'$ is dominant (in other words $K$ is an extension of the function field of $S'$)
and there exists an integer $d | n$ such that $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ is nonzero.
Conversely, nonvanishing of $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ in the lemma implies $BC(f, n, q)$ isn't true as Lemma 59.80.5 shows that $R^ q(\mathop{\mathrm{Spec}}(K) \to S')_*\mathbf{Z}/d\mathbf{Z} = 0$.
Proof.
First choose a diagram and $\mathcal{F}$ as in Lemma 59.88.6. We may and do assume $S'$ is affine (this is obvious, but see proof of the lemma in case of doubt). By Lemma 59.88.3 we may assume $K$ is algebraically closed. Then $\mathcal{F}$ corresponds to a $\mathbf{Z}/n\mathbf{Z}$-module. Such a modules is a direct sum of copies of $\mathbf{Z}/d\mathbf{Z}$ for varying $d | n$ hence we may assume $\mathcal{F}$ is constant with value $\mathbf{Z}/d\mathbf{Z}$. By Lemma 59.88.4 we may replace $S'$ by the normalization of $S'$ in $\mathop{\mathrm{Spec}}(K)$ which finishes the proof.
$\square$
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